
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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The Notch mutation is a deletion on the X chromosome of Drosophila melanogaster. Female flies heterozygous for Notch have an indentation on the margins of their wings; Notch is lethal in the homozygous and hemizygous conditions. The Notch deletion covers the region of the X chromosome that contains the locus for white eyes, an X-linked recessive trait. Give the
Q.A white-eyed Notch female is mated with a red-eyed male.
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- Red eyes (r), brown color (b), and curled wings (c) are 3 recessive mutations that occur in house flies. These genes have already been mapped: r-b = 12 mu b-c = 18 mu r-c = 30 mu A fly with red eyes, brown color, and curled wings is crossed with a fly homozygous for the wild-type traits. The F1 males are crossed with females with the three mutant traits and 2000 progeny are produced. Find the number of observed double crossovers if the interference is 0.38. O 43 O 62 O 27 O 16arrow_forwardIn Drosophila, the X-linked recessive mutation vermilion (v) causes bright red eyes, in contrast to the brick-red eyes of wild type. A separate autosomal recessive mutation, suppressor of vermilion (su-v), causes flies homozygous or hemizygous for v to have wildtype eyes. In the absence of vermilion alleles, su-v has no effect on eye color. Determine the F1 and F2 phenotypic ratios from a cross between a female with wild-type alleles at the vermilion locus, but who is homozygous for su-v, with a vermilion male who has wildtype alleles at the su-v locusarrow_forwardIn Drosophila, the genes st (scarlet eyes), ss (spineless bris- tles), and e (ebony body) are located on chromosome 3, with map positions as indicated: st SS e 44 58 70 Each of these mutations is recessive to its wild-type allele (st*, dark red eyes; ss*, smooth bristles; e*, gray body). Phenotypically wild-type females with the genotype st ss e*/st* st* ss+ e were crossed with triply recessive males. Predict the phenotypes of the progeny and the frequen- cies with which they will occur assuming (a) no interfer- ence and (b) complete interference.arrow_forward
- XX individuals can be sex-reversed males if one of their X chromosomes contains a translocated portion of the Y chromosome that includes the SRY gene. In light of what you now know about X-chromosome inactivation, it seems that this karyotype might have more complex consequences. In fact, although most such males are completely sex-reversed, X-chromosome inactivation can cause some of these XX males to have varying degrees of residual female characteristics. a. X-chromosome inactivation in normal XX embryos occurs earlier in development than SRY production in normal XY embryos. When present on a translocation X chromosome, SRY is subject to inactivation. Formulate a hypothesis to explain why many XX individuals with a translocation X chromosome that includes the SRY gene are not completely sex-reversed (male). b. Based on your answer to part (a), why do you think some individuals with this karyotype are completely sex-reversed?arrow_forwardThere are two genetic disorders that result from mutation in imprinted genes: Prader-Willi syndrome, Angelman syndrome. Angelman syndrome results from deletion of UBE3A, which is a gene imprinted such that only the maternal copy is expressed. In the pedigree above, individual I-1 is heterozygous for a deletion of UBE3A and does not have Angelman syndrome. Individual I-2 is homozygous wild type for UBE3A. Which individuals in the pedigree are at risk for exhibiting Angelman syndrome, if any? (Who could potentially have the syndrome, based on what alleles it is possible for them to inherit and express?) Question 8 options: Only I-1 could have been at risk. If he does not have the syndrome, no one in the pedigree could. Only III-1 is at risk I-1, II-2, and III-1 are all at risk Only II-2 is at risk No one in the pedigree is at risk Both II-2 and III-1 are at…arrow_forwardFemales of wild-type Strain A and males of mutant Strain B, as well as females of mutant Strain B and males of wild-type Strain A, make reciprocal crosses. Explain why reciprocal crosses are needed in genetics experiments involving Drosophila fruit flies.arrow_forward
- Cinnabar eyes (cn) and reduced bristles (rd) are linked autosomal recesive characters in Drosophila fruit flies. A homozygous wild-type female was crossed to a reduced, cinnabar male, and the F1 females were then crossed with cinnabar reduced males to obtain the F2. Of the 200 F2 offspring obtained, 86 were wild type, 13 were cinnabar, 17 were reduced and 84 were reduced and cinnabar.. What is the map distance between the cn and rd alleles? 8,4 mu 15 mu 17 mu 30 mu 13 muarrow_forwardPlease explain how you do this question step by step I am very confused! thank you:) You have three independent mutant alleles in the Drosophila gene no-antenna: nan1, nan2 and nan3. You determine the phenotype of Drosophila that are heterozygous for the three alleles (heterozygous for a wild-type allele and a mutant allele), and that are homozygous for the three mutant alleles. The antenna is composed of three segments that are followed at the distal end by a feathery arista (that is the antenna is composed of three segments and an arista). Allele nan1 nan2 nan3 heterozygous Wild-type No arista Wild-type homozygous No arista No antenna No antenna nan1 is a __x__ allele, nan2 is a __y__ allele, and nan3 is a __z__ allele. X Y Z A Dominant negative Null Hypomorphic B Null Dominant negative Hypomorphic C Null Hypomorphic Dominant negative D Hypomorphic Dominant negative Null E Dominant negative Hypomorphic Null Referencing the table above, select the…arrow_forward
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